Chapter 13, Problem 69P

A string on a musical instrument is held under tension T and extends from the point x = 0 to the point x = L. The string is overwound with wire in such a way that its mass per unit length μ(x) increases uniformly from μ_o at x = 0 to μ_L at x = L. (a) Find an expression for μ(x) as a function of x over the range 0 ≤ x ≤ L. (b) Find an expression for the time interval required for a transverse pulse to travel the length of the string.

To determine

The expression for $\mu \left(x\right)$ as a function of $x$ over the range $0\le x\le L$.

Answer to Problem 69P

The expression for $\mu \left(x\right)$ as a function of $x$ over the range $0\le x\le L$ is $\mu \left(x\right)=\left(\frac{{\mu }_{\text{L}}-{\mu }_0}{L}\right)x+{\mu }_0$.

Explanation of Solution

Given info: The tension in the string is $T$ and the extension point for string is $0$ to $L$.

Assume the linear expression for the linear density $\mu \left(x\right)$ is,

$\mu \left(x\right)=ax+b$ (1)

Here,

$\mu \left(x\right)$ is the linear density at $x$.

Substitute $0$ for $x$ in equation (1).

$\begin{array}{c}{\mu }_0=a\times 0+b\\ =b\end{array}$

Substitute $L$ for $x$ in equation (1).

${\mu }_{\text{L}}=aL+b$ (2)

Substitute $\mu \left(0\right)$ for $b$ in equation (2).

$\begin{array}{c}{\mu }_{\text{L}}=aL+{\mu }_0\\ a=\frac{{\mu }_{\text{L}}-{\mu }_{\text{0}}}{L}\end{array}$

Substitute $\frac{{\mu }_{\text{L}}-{\mu }_0}{L}$ for $a$ and ${\mu }_0$ for $b$ in equation (1).

$\mu \left(x\right)=\left(\frac{{\mu }_{\text{L}}-{\mu }_0}{L}\right)x+{\mu }_0$

Conclusion:

Therefore, the expression for $\mu \left(x\right)$ as a function of $x$ over the range $0\le x\le L$ is $\mu \left(x\right)=\left(\frac{{\mu }_{\text{L}}-{\mu }_0}{L}\right)x+{\mu }_0$.

To determine

The expression for the time interval required for transverse pulse to travel the length of the string.

Answer to Problem 69P

The expression for the time interval required for transverse pulse to travel the length of the string is $\Delta t=\frac{2L}{3\left({\mu }_L-{\mu }_0\right)\sqrt{T}}\left[{\mu }_L{}^{\frac{3}{2}}-{\mu }_0{}^{\frac{3}{2}}\right]$.

Explanation of Solution

Given info: The tension in the string is $T$ and the extension point for string is $0$ to $L$.

Formula to calculate the speed of the wave is,

$v=\frac{dx}{dt}=\sqrt{\frac{T}{\mu \left(x\right)}}$

Here,

$v$ is the speed of the wave.

$T$ is the tension in the string.

Rearrange the above equation.

$dx\sqrt{\mu \left(x\right)}=\sqrt{T}dt$ (3)

Integrate the right hand side of the equation (3) from $0$ to $L$ and right hand side from $0$ to $t_{\text{final}}$.

${\displaystyle \underset{0}{\overset{L}{\int }}dx\sqrt{\mu \left(x\right)}}={\displaystyle \underset{0}{\overset{t_{\text{final}}}{\int }}\sqrt{T}dt}$ (4)

Substitute $ax+b$ for $\mu \left(x\right)$ in equation (4).

${\displaystyle \underset{0}{\overset{L}{\int }}dx\sqrt{ax+b}}={\displaystyle \underset{0}{\overset{t_{\text{final}}}{\int }}\sqrt{T}dt}$ (5)

Assume,

$n=ax+b$

When $x=0$ then $n=b$ and when $x=L$ then $n=ax+b$.

Substitute $ax+b$ for $n$ in equation (5).

${\displaystyle \underset{0}{\overset{L}{\int }}\sqrt{n}dx}={\displaystyle \underset{0}{\overset{t_{\text{final}}}{\int }}\sqrt{T}dt}$ (6)

Integrate the left hand side of the equation (6) from $b$ to $ax+b$.

$\begin{array}{c}{\displaystyle \underset{b}{\overset{ax+b}{\int }}dx\sqrt{ax+b}}={\displaystyle \underset{0}{\overset{t_{\text{final}}}{\int }}\sqrt{T}dt}\\ \frac{2}{3a\sqrt{T}}\left[{\left(aL+b\right)}^{\frac{3}{2}}-b^{\frac{3}{2}}\right]=t_{\text{final}}\end{array}$ (7)

Substitute $\mu \left(L\right)$ for $aL+b$, $\mu \left(0\right)$ for $b$ and $\frac{\mu \left(L\right)-\mu \left(0\right)}{L}$ for $a$ in equation (7).

$\frac{2}{3\frac{\mu \left(L\right)-\mu \left(0\right)}{L}\sqrt{T}}\left[\mu {\left(L\right)}^{\frac{3}{2}}-\mu {\left(L\right)}^{\frac{3}{2}}\right]=t_{\text{final}}$

This can be written as,

$\begin{array}{c}\frac{2L}{3\left({\mu }_L-{\mu }_0\right)\sqrt{T}}\left[{\mu }_L{}^{\frac{3}{2}}-{\mu }_0{}^{\frac{3}{2}}\right]=t_{\text{final}}=\Delta t\\ \Delta t=\frac{2L}{3\left({\mu }_L-{\mu }_0\right)\sqrt{T}}\left[{\mu }_L{}^{\frac{3}{2}}-{\mu }_0{}^{\frac{3}{2}}\right]\end{array}$

Conclusion:

Therefore, the expression for the time interval required for transverse pulse to travel the length of the string is $\Delta t=\frac{2L}{3\left({\mu }_L-{\mu }_0\right)\sqrt{T}}\left[{\mu }_L{}^{\frac{3}{2}}-{\mu }_0{}^{\frac{3}{2}}\right]$.